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# answer%20HW5 - HW 5 Solutions Problem 1 Taking the z axis...

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HW 5 Solutions Problem 1 Taking the z - axis to be the direction of the electric field and using z = r cos θ , the only non-zero transition matrix element is between the ground state and the 2 p state with m l = 0. Hence, H 210 , 100 ( t ) = - eE 0 τ 2 + t 2 drψ * 210 ( r ) r cos θψ 100 ( r ) = - eE 0 τ 2 + t 2 1 4 2 πa 3 0 2 π 0 π 0 cos 2 θ sin θ 0 dr r 3 r a 0 exp[ - 3 r/ 2 a 0 ] = - eE 0 a 0 τ 2 + t 2 2 7 2 3 5 (1) The transition probability is thus P = e 2 E 2 0 a 2 0 ¯ h 2 2 15 3 10 -∞ dt exp( - ωt ) τ 2 + t 2 2 = π 2 e 2 E 2 0 a 2 0 ¯ h 2 τ 2 2 15 3 10 exp( - 2 ωτ ) (2) where ω = 3 me 4 8(4 π ¯ h ) 2 (3) Problem 2 We denote the vacuum state by | 0 and excited states by | n . These are the usual energy eigenkets of the 1D SHO. We note that n | x 2 | 0 = n ¯ h 2 0 a 2 - + a + a - + a - a + + a 2 + 0 = ¯ h 2 0 δ n 0 + 2 δ n 2 (4) where ω 0 is the classical angular frequency. Using (1), and to first-order in time-dependent pertur- bation theory, c (1) n ( t ) = - i ¯ h t 0 e inω 0 t n | H | 0 dt = - i ¯ h K t 0 e inω 0 t e - t /τ ¯ h 2 0 δ n 0 + 2 δ n 2 dt (5) Thus, according to first-order theory, it can only transit to n = 2. c (1) 2 ( t ) = - i 2 K 2 0 t 0 e 2 0 t - t /τ dt = - iK 2 2 0 e (2 0 - 1 ) t - 1 2 0 - 1 τ (6) For t τ | c (1) 2 ( t ) | 2 K 2 2 m 2 ω 2 0 1 4 ω 2 0 + 1 2 (7) 1

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Problem 3 The Hamiltonian reads: H = β [ S 1 x S 2 x + S 1 y S 2 y + S 1 z S 2 z ] . (8) In the basis set of | + + , | - - , | + - , | - + , the matrix representation of the Hamiltonian H ij = i | H | j is H = β ¯ h 2 4 1 0 0 0 0 1 0 0 0 0 - 1 2 0 0 2 - 1 (9) Since H is the time-evolution operator, we have from (9): i ¯ h∂ t | + + = β ¯ h 2 4 | + + (10) i ¯ h∂ t | - - = β ¯ h 2 4 | - -
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